How do we find an isomorphism between P_4 to P*_4 (dual space of P_4)

Question

I obviously know that there is an isomorphism between a vector space and its dual space since their dimensions are equal, but I have problem in explicitly showing an isomorphism between them(In this case, its polynomials) May I get some helps...?

Answer 1

Choose a basis $(a,b,c,d,e)$ of $P_4$ and a basis $(f,g,h,i,j)$ of its dual. Then take the only linear map $F$ such that $F(a)=f,F(b)=g$ and so on.

Answer 2

Lagrange interpolation gives an explicit isomorphism between $P_n$ and $P_n^*$: choose $n+1$ different scalars $x_0,x_1,\dots,x_n$ and send $p \in P_n$ to $p(x_0)E_0+\cdots+ p(x_n)E_n \in P_n^*$, where $E_j: p \mapsto p(x_j)$ is an evaluation functional.

Under this isomorphism, the basis dual to $E_j$ is the Lagrange basis: $$ L_j(x) = \prod_{i\neq j} \frac{x-x_i}{x_j-x_i} $$

Answer 3

We can think of $\mathbf R^n$ as its own dual space using the standard dot product: the bilinear pairing $\mathbf R^n \times \mathbf R^n \rightarrow \mathbf R$ where $(v,w) \mapsto v \cdot w$ lets us view $v \in \mathbf R^n$ as an element of $(\mathbf R)^n$ by $\varphi_v(w) = v \cdot w$ as $w$ varies. Then $v \mapsto \varphi_v$ is $\mathbf R$-linear from $\mathbf R^n$ to its dual space and is injective, so it is an isomorphism since $\mathbf R^n$ and its dual space have the same (finite) dimension.

If you are working over the real numbers then the dual space of $P_n$ can be thought of in terms of derivatives acting on $P_n$. Let $D = d/dx$, which acts linearly on $P_n$. For $f \in P_n$ let $f(D)$ be the corresponding polynomial in $D$: if $f(x) = c_0 + c_1x + \cdots + c_nx^n$ then $f(D) = c_0 + c_1D + \cdots + c_nD^n$, so $f(D)(g) = c_0g(x) + c_1g'(x) + \cdots + c_ng^{(n)}(x)$. This is $\mathbf R$-linear in $g$ when $f$ is fixed, and it is $\mathbf R$-linear in $f$ when $g$ is fixed, so by composing with evaluation at $0$ we get a bilinear pairing $P_n \times P_n \rightarrow \mathbf R$ by $(f,g) \mapsto (f(D)g)(0)$.

For example, if $f(x) = x^i$ and $g(x) = x^j$ then $f(D)g = D^i(x^j) = j(j-1)\cdots (j-(i-1))x^{j-i}$. When $i>j$ the exponent $j-i$ is negative, so $x^{j-i}$ is not a polynomial, but also the coefficient is $0$, so when $i > j$ we have $D^i(x^j) = 0$ in $P_n$, and thus $(D^i(x^j))(0) = 0$ in $\mathbf R$ when $i > j$. We also have $(D^i(x^j))(0) = 0$ for $i < j$ since $x^{j-i}$ vanishes at $x = 0$ when $i < j$, and for $i = j$ we have $D^i(x^j) = i!$ is a nonzero constant, so $(D^i(x^i))(0) = i!$. Thus $(x^i,x^j) = 0$ if $i \not= j$ and $(x^i,x^i) = i! \not= 0$. This shows us that $(x^i,x^j) = (x^j,x^i)$ for all $i$ and $j$ up to $n$, so from bilinearity $(f,g) = (g,f)$ for all $f$ and $g$ in $P_n$, which a priori is not so obvious. An explicit formula is $(\sum_{i=0}^n a_ix^i,\sum_{j=0}^n b_jx^j) = \sum_{i=0}^n a_ib_ii!$, so $(f,f) = \sum_{i=0}^n a_i^2i! \geq 0$, with $(f,f) = 0$ only when each coefficient of $f$ is $0$. Hence our pairing on $P_n$ is an inner product (positive-definite symmetric bilinear pairing), and we can think of $P_n$ as its own dual space by associating to each $f \in P_n$ the functional $\varphi_f \colon P_n\rightarrow \mathbf R$ where $\varphi_f(g) = (f,g)$ as $g$ varies.

I learned of this construction from the chapter on harmonic and symmetric polynomials in Lang's Math Talks for Undergraduates, where he does the same construction on polynomials of degree up to $n$ in several variables, not just one variable (using partial derivatives).